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Dear Moni
The equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0. Represents a second degree equation where a, h, b doesn’t variables simultaneously. Let a ≠ 0. Now, the above equation becomes a2 x2 + 2ax (hy + g) = aby2 – 2afy – ac on completing the square on the left side, we get, a2 x2 + 2ax (hy + g) = y2 (h2 – ab) + 2y (gh – af) + g2 – ac. i.e. (ax + hy + g) = + √y2(h2–ab)+2y(gh–af) +(g2–ac) We cannot obtain x in terms of y, involving only terms of the first degree, unless the quantity under the radical sign be a perfect square. The condition for this is, (gh – af)2 = (h2 – ab) (g2 – ac) i.e. g2h2 – 2afgh + a2f2 = g2h2 – abg2 – abg2 – ach2 + a2bc cancelling and diving by a, we have the required condition abc + 2fgh – af2 – af2 – bg2 – ch2 = 0 Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it hereand we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best.Regards,Askiitians ExpertsBadiuddin
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